a) 16x²-(x²+4)²= (4x)²-(x²+4)²

                      = (4x-x²-4)(4x+x²+4)

\(\text{b) 27x^3-54x^2+36x-8=[(3x)^3-3.(3x)^2.2+3.3x.2^2-2^3}]\)

                                                      = (3x-2)³
\(\text{c) (x+y)^3 - (x-y)^3= (x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]}\)

                                   =2y(x²+2xy+y²+x²-y²+x²-2xy+y²)

                                   = 2y(3x²+y²)

a) = a( a^2-1 )

= a(a-1)(a+1)

b) =16x^3- 16x^2 +4x^2-4x+7x -7

=16x^2(x-1) +4x(x-1) +7(x-1)

=(x-1)(16x^2 +4x+7)

bạn có thể viết rõ ràng hơn đc k ? mình k hiểu đề bài cho lắm!!!

a) a^3-a

b)16x^3-12x^2+3x-7

phân tích thành nhân tử

dat y^2+y=z cho gon

\(z^2-9z+20=z^2-4z-5z+20=z\left(z-4\right)-5\left(z-4\right)=\left(z-4\right)\left(z-5\right)\)

\(thaylai:\left(y^2+y-4\right)\left(y^2+y-5\right)\)

thay hay thi "k" ko hay thi thoi 

x^3 - x + 3x^2y + 3xy^2 + y^3 - y

=x³+y³+3x²y+3xy²-x-y

=(x+y)(x²-xy+y²)+3xy(x+y)-(x+y)

=(x+y)(x²-xy+y²+3xy-1)

=(x+y)(x²+2xy+y²-1)

=(x+y)[(x+y)²-1]

=(x+y)(x+y-1)(x+y+1)

x^2 + 5x - 6

=x²-x+6x-6

=x.(x-1)+6.(x-1)

=(x-1)(x+6)

4x² - 36x + 56

= 4x² - 28x - 8x + 56

= ( 4x² - 28x ) - ( 8x - 56 )

= 4x( x - 7 ) - 8( x - 7 )

= ( x - 7 )( 4x - 8 )

= 4( x - 7 )( x - 2 )

x³-x+3x²y+3xy²+y³-y

=x²(x-1)+3(x²y+xy²)+y²(y-1)

=x²(x-1)+3(x².y+y².x)+y²(y-1)

=x²(x-1)+3{[x(x+1)+y(y+1)]}+y²(y-1)

=x²(x-1)+3.x(x+1)+3.y(y+1)+y²(y-1)

=x²(x-1)+2x²+3.x(x+1)+3.y(y+1)+y²(y-1)+2y²-2x²-2y²

=x²(x+1)+3.x(x+1)+3.y(y+1)+y²(y+1)-2x²-2y²

=(x²+3)(x+1)+(y²+3)(y+1)-2(x²+y²)

ta có : (x*3+3x*2y+3xy*2+y*3)-(x+y)

=(x+y)*3-(x+y)

=(x+y)((X+Y)*2-1)

(x+y)(x+y+1)(x+Y-1)

\(x^3-x+3x^2y+3xy^2+y^3-y=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)

=\(\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\cdot\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\cdot\left(x+y+1\right)\cdot\left(x+y-1\right)\)

\(49.\left(y-4\right)^2-9y^2-36y-36\)

\(=7^2\left(y-4\right)^2-\left(9y^2+36y+36\right)\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28+3y+6\right).\left(7y-28-3y-6\right)\)

\(=\left(10y-22\right).\left(4y-34\right)\)

\(=4.\left(5y-11\right).\left(2y-17\right)\)

\(49\left(y-4\right)^2-9y^2-36y-36\)

\(=\) \(4\left(2y-17\right)\left(5y-11\right)\)